I wanted to know what is the Feynman rules for current-current vertex like this one:
$$ {\cal{ L}} = G^\prime_F \hspace{2mm} \bar{d} \gamma^\mu (1-\gamma^5) u\hspace{3mm} \bar{s}\gamma_\mu (1-\gamma^5)c $$
And what is the simplest way to derive it ? I was trying the following way:
For example in QED, we have $$ {\cal{L}}_{int} = e \bar{\psi}\gamma^\mu \psi A_\mu $$ and omitting the fields (taking partial derivatives) I achieve the vertex:
$$ V_{qed} = i e \gamma^\mu $$ For the current-current case, I wanted to apply the same method, the the Feynman vertex I achieve is this:
$$ V_{current-current} = i G^\prime _F \gamma^\mu (1-\gamma^5) \gamma_\mu (1-\gamma^5) $$ But this appear to be meaningless, because if I simplify this vertex using the anticommutation $ \{\gamma^\mu, \gamma^5 \} = 0 $ and the properties $ (\gamma^5)^2 = 1_{4\times 4} $ and $\gamma^\mu \gamma_\mu = 4 1_{4\times 4} $, then we can see that:
$$ V_{current-current} = i G^\prime _F \gamma^\mu (1-\gamma^5) \gamma_\mu (1-\gamma^5) $$ $$ = i G^\prime _F (1+\gamma^5) \gamma^\mu \gamma_\mu (1-\gamma^5) $$ $$ = 4 i G^\prime _F (1+\gamma^5) (1-\gamma^5) $$ $$ = 4 i G^\prime _F (1 - (\gamma^5)^2) $$ $$ = 4 i G^\prime _F (1 - 1) = 0 $$
What's going on here ? Can anyone please help me find the correct vertex rule !